Question: $h(n) = -4n^{3}+3n^{2}+3n+f(n)$ $f(t) = -t^{2}+6t$ $ f(h(1)) = {?} $
Explanation: First, let's solve for the value of the inner function, $h(1)$ . Then we'll know what to plug into the outer function. $h(1) = -4(1^{3})+3(1^{2})+(3)(1)+f(1)$ To solve for the value of $h$ , we need to solve for the value of $f(1)$ $f(1) = -1^{2}+(6)(1)$ $f(1) = 5$ That means $h(1) = -4(1^{3})+3(1^{2})+(3)(1)+5$ $h(1) = 7$ Now we know that $h(1) = 7$ . Let's solve for $f(h(1))$ , which is $f(7)$ $f(7) = -7^{2}+(6)(7)$ $f(7) = -7$